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2 years ago

When a base reacts with an acid, the acid/base reaction always producess

Chemistry

1 answer:

Lelechka [254]2 years ago

3 0

Explanation:

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water

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13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What

rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

of MgCO3.3H2O in the mixture?

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

Mass water:

3.883g - 2.927g = 0.956g water

Moles water -18.01g/mol-

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

Moles MgCO3.3H2O:

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

Mass percent:

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

6 0

2 years ago

Is facilitated diffusion passive or active transport?

bazaltina [42]

Answer:

Its passive

Explanation:

Facilitated diffusion is a type of passive transport that allows substances to cross membranes with the assistance of special transport proteins.

6 0

2 years ago

Jack lives in a small city in the Pacific Northwest, where the temperatures are cool all year and the sky is frequently cloudy a

kirill [66]

Answer:

Preocouparcor ,imsm oilp

Explanation:

3 0

2 years ago

A sample of argon gas collected at a pressure of 0.918 atm and a temperature of 279 K is found to occupy a volume of 719 millili

babymother [125]

Yeah i have 0 idea i’m sorry.

3 0

2 years ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

3 years ago